CBSE NCERT Solutions for Class 12 Chemistry Chapter 2 · Class- XII-CBSE-Chemistry Solutions Practice more on Solutions Page - 1 CBSE NCERT Solutions for Class 12 Chemistry Chapter

Class- XII-CBSE-Chemistry Solutions

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CBSE NCERT Solutions for Class 12 Chemistry Chapter 2

Back of Chapter Questions

2.1. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4)

if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution:

Given: mass of benzene (C6H6 solute⁄ ) = 22 gram

Mass of carbon tetrachloride (CCl4 solvent⁄ ) = 122 gram

We Find: mass percentage of benzene (C6H6) = ?

Mass percentage of carbon tetrachloride (CCl4) = ?

Formula required: Mass percentage of solute =Mass of solute (gm)

Total mass of the solution(gm)×

100%

=Mass of solute (gm)

mass of the solute (gm) + mass of the solvent (gm)× 100%

Calculation: Mass percentage of C6H6 =Mass of C6H6(gm)

Total mass of the solution(gm)× 100%

=Mass of C6H6 (gm)

mass of the solute (gm) + mass of the solvent (gm)× 100%

=Mass of C6H6(gm)

Mass of C6H6(gm) + Mass of CCl4× 100%

=22

22 + 122× 100%

= 15.28%

Mass percentage of CCl4 =Mass of CCl4

Total mass of the solution× 100%

=Mass of CCl4(gm)

Mass of C6H6(gm) + Mass of CCl4(gm)× 100%

=122

22 + 122× 100%

= 84.72%

Alternatively,

Mass percentage of CCl4 = (100 − 15.28)% = 84.72%

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2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in

carbon tetrachloride.

Solution:

Give % by mass C6H6 in CCl4 = 30%

benzene in solution containing 30% by mass in carbon tetrachloride.

Assume:

Let the total mass of the solution be 100 g and the mass of benzene be 30

∴ Mass of carbon tetrachloride = (100 − 30)g = 70 g

We know that Molar mass of benzene (C6H6) = (6 × 12 + 6 × l)g mol−1 =

78 g mol−1

Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 355 = 154 g mol−1

The formula used for mole fraction:

Number of moles of solute (n1)

Number of moles of solute (n1) + Number of moles of solute(n2)

∴ Number of moles of C6H6(n1) =30

78mol = 0.3846 mol

∴ Number of moles of CCl4(n2) =70

154 mol = 0.4545 mol

the mole fraction of C6H6 is given as:

Number of moles of C6H6(n1)

Number of moles of C6H6(n1) + Number of moles of CCl4(n2)

=0.3846

0.3846 + 0.4545

= 0.458

2.3. Calculate the molarity of each of the following solutions:

(a). 30 g of Co(NO3)2. 6H2O in 4.3 L of solution

(b). 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Solution:

Molarity is given by:

(a) Given : Co(NO3)2. 6H2O = 30 gm

Volume of solution = 4.3 liter




We know : Molar mass of Co(NO3)2. 6H2O = 59 + 2(14 + 3 × 16) +

6 × 18

= 291 g mol−1

Moles of (NO3)2. 6H2O =(weight)(gm)

molecularweight =

30

291mol

= 0.103 mol

The formula used : Molarity =Moles of solute

Volume of solution in litre

Therefore, molarity=0.103 mol

4.3 L= 0.023 M

(b) Given : 30 mL of 0.5 M H2SO4

Initial Volume of H2SO4 solution =30 ml

Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

Number of moles present in 30 mL of 0.5 M H2SO4 =0.5×30

1000 mol

= 0.015 mol

Therefore,Molarity =Moles of solute

Volume of solution in litre=

Moles of solute×1000

Volume of solution in Mili litre

Moles of solute = MV(litre)

Final Volume of the solution after dilution = molarity =MiVi

Vf=

0.015

0.5 Lmol

= 0.03 M

2.4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal

aqueous solution.

Solution:

Given : molality of aqueous solution of urea = 0.25

we know

weight of solution = 2.5 kg = 2500 g

Molar mass of urea (NH2CONH2) = 2(1 x 14 + 2 x 1) + 1 x 12 + 1 x 16 =

60 g mol−1

0.25 molar aqueous solution of urea means that

1000 g of water contains 0.25 mol = (0.25 x 60)g of urea

= 15 g of urea




That is,

Weight of solute = 15 g

Weight of solvent = 1000 g, weight of solution = (15 + 1000)g

(1000 + 15) g of solution contains 15 g of urea = 15 × 2500

Therefore, 2.5 kg (2500 g) of solution contains =15×2500

1000+15 = 36.945 g

= 36.945 g of urea (approximately)

Hence, the mass of urea required = 36.945 g

2.5. Calculate

(a) molality

(b) molarity and

(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is

1.202 g mL−1.

Solution:

Given density of aqueous KI solution =1.202

gm

mol

% mass by mass of KI = 20%

(a) The molar mass of KI = 39 + 127 = 166 g mol−1

That is,

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in

100 g of solution.

20 g Of KI is present in (100 − 20)g of water = 80 g of water

Therefore, molality of the solution (m) =Moles of KI

Mass of water in kg

Moles of KI = wt/mol =20

166mole

Weight of water (solvent) = 80 g =80

100kg.

=

201660.08

molal

= 1.506 molal

% mass by mass = 20%




Mass by mass means

Mass of solution = 100 g and mass of solvent = 20g

(b) given: the density of the solution = 1.202 g mL−1

Volume of 100 g solution =mass

density

=100 g

1.202 g mL−1

= 83.19 mL

= 83.19 × 10−3 L

Moles of solute = weight/molecular weight =20

166

Formulae used molarity (M) =moles of solute

volume of solution (litre)

Therefore, the molarity of the solution =20

166mol

83.19×10−3L

= 1.45 M

Given 20% w/w KI


Weight of solution = 100 g

Weight of solvent = weight of solution – weight of solute

⇒ weight of solvent = 100 − 20 = 80 g

(c) Moles of KI =weight of KI

Molecular weight of KI

=20

166= 0.12 mol

Moles of water =weight of H2Ogm

molecular weight of H2O

80

18= 4.44 mol

Exercises

2.1. Define the term solution. How many types of solutions are formed? Write briefly

about each type with an example.

Solution:

A solution is a homogeneous mixture of two or more than two components (solute

and solvent)




There are three types of solutions, they are:

(i) Gaseous solution:

A gaseous solution is a solution in which the solvent is gas is called a

gaseous solution. In these solutions, the solute may be liquid, solid, or gas.

For example, a homogenous mixture of oxygen and hydrogen gas is a

gaseous solution.

(ii) Liquid solution:

A liquid solution is a solution in which the solvent is a liquid is known as

a liquid solution. The solute in these solutions may be gas, liquid, or solid.

The mixture of milk and water is an example of a liquid solution.

(iii) Solid solution:

A solid solution is a solution in which the solvent is a solid dissolved. The

solute may be gas, liquid or solid. Metal alloys are examples of solid

solutions.

2.2. Give an example of a solid solution in which the solute is a gas.

Solution:

A solid solution is formed between two substances solute and solvent (where one

substance has very large particles and the other is having very small particles), an

interstitial solid solution will be formed. Consider an example, a solution of

hydrogen in palladium is a solid solution in which the solute is a gas (H2) and the

solvent is Pd.

2.3. Define the following terms:

(i) Mole fraction

(ii) Molality

(iii) Molarity

(iv) Mass percentage.

Solution:

(i) Mole fraction:

The mole fraction of a component in a mixture is defined as the ratio of

the number of moles of the component to the total number of moles of all

the components in the mixture.

i.e., Mole fraction of component (X) =Number of moles of the component

Total number of moles of all components




If, in a binary solution, the number of moles of the solute is nA and of the

solvent is nB, then the mole fraction of the solute in the solution is (xA) =nA

nA+nB

Similarly, the mole fraction of the solvent in the solution is (xB) =nB

nA+nB

And, moles fraction of solute (xA) +mole fractions of the solvent (xB) =1

Molarity is a unitless quantity.

(ii) Molality

It is unitless quantity

The number of moles of the solute per kilogram of the solvent is called as

molality. It is expressed as:

Molality (m)=Moles of solute (n)

Mass of solvent in (kg)=

moles of solute × 1000

mass of solvent (g)

Unit is molal

(iii) Molarity

The number of moles of the solute dissolved in one litre of the solution is

called molarity of the solution.

It is expressed as:

Molarity (M) =Moles of solute

Volume of solution in (litre)=

moles of solute (1000)

volume of solution (ml)

(iv) Mass percentage:

mass of solute

mass of solution× 100 =

mass of solute

mass of solute + mass of solvent× 100

2.4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in

aqueous solution. What should be the molarity of such a sample of the acid if the

density of the solution is 1.504 g mL−1?

Solution:

Given density of solution = 1.50 g/mL

We know that concentrated nitric acid used in laboratory work is 68% nitric acid

by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in

100 g of the solution.

Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1




The formula used: M =moles of solute

volume of solution(litre)

Then, the number of moles of HNO3 =weight

molecular weight=

68

63 mol

= 1.079 mol

We know that density of solution = 1.504 g mL−1

i.e., the mass of solution = 100 g by mass percentage

∴ Volume of 100 g solution =mass

density =

100

1.504 mL

= 66.49 mL

= 66.49 × 10−3 L

∴ 1 ml = 10−3litre

Molarity of solution =1.079 mol

66.49×10−3L

= 16.23 M

68% Nitric acid by mass means that in a solution of 100g, 68g is solute and rest

is solvent i. e. 32g.

2.5. A solution of glucose in water is labelled as 10% w/w, what would be the

molality and mole fraction of each component in the solution? If the density of the

solution is 1.2 g mL−1, then what shall be the molarity of the solution?

Solution:

Given,

10% w/w solution of glucose in water means that 10 g of glucose is present in

100 g of the solution i. e. , 10 g of glucose is present in (100 − 10)g = 90 g of

water.

We know, Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 =

180 g mol−1

Then, the number of moles of glucose= weight

molecular weight =

10

180 mol

= 0.056 mol

∴ The molality of solution =moles of volume

weight of solvent (kg)=

0.056 mol

0.09kg= 0.62 molal

Number of moles of water =90 g

18 g mol−1 = 5 mol




Mole fraction of glucose(xg) =moles og glucose

moles of glucose + moles of H2O =

0.056

0.056+5

⇒ Mole fraction of glucose = 0.011

And, the mole fraction of water (xw) = 1 − moles of glucose = 1 − xg

= 1 − 0.011

= 0.989

If the density of the solution is 1.2 g mL−1, then the volume of the 100 g solution

=mass

density

Volume of solution (v) =100g

1.2g mL1= 83.33 mL = 83.33 × 10−3L

∴ Molarity of the solution =moles of solute

volume of solution (litre)=

0.056mol

83.33×10−3L

= 0.67 M

2.6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of

Na2CO3 and NaHCO3 containing equimolar amounts of both?

Solution:

Given molarity of HCl = 0.1 M

Volume solution =?

Assume the amount of Na2CO3 in the mixture to be x g.

Then, the amount of NaHCO3 in the mixture is (1 − x) g.

Weight of mixture of Na2CO3 & NaHCO3 = 1 gm

Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16 = 106 g mol−1

∴ Number of moles of Na2CO3 weight Na2CO3

molecular weight of Na2CO3=

x

106 mol

Molar mass of NaHCO3 = 1 × 23 × 1 × 1 × 12 + 3 × 16 = 84 g mol−1

∴ Number of moles of NaHCO3weight NaHCO3

molecular weight of NaHCO3=

1−x

84 mol

According to question Na2CO3 and NaHCO3 have equimolar so moles of

Na2CO3 = moles of NaHCO3

x

106=

1 − x

84

⇒ 84x = 106 − 106x




⇒ 190x = 106

x = 0.5579

Therefore, the number of moles of Na2CO3 =x

106=

0.5579

106 mol

= 0.0053 mol

And the number of moles of NaHCO3 =1−x

84=

1−0.5579

84

= 0.0053 mol

HCl Reacts with Na2CO3 and NaHCO3 according to the following equation.

2HCl + Na2CO3 ⟶ 2NaCl + H2O + CO2 … (1)

2 mol 1 mol

HCl + NaHCO3 ⟶ NaCl + H2O + CO2 … (2)

1 mol 1 mol

1 mol Of Na2CO3 reacts with 2 mol of HCl.

Therefore, 0.0053 mol of Na2CO3 reacts with 2 × 0.0053 mol = 0.0106 mol.

Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl. Then for mole of HCl =

moles of NaHCO3 = 0.0053

Total moles of HCl required = (0.0106 + 0.0053) mol = 0.0159 mol

⇒ 0.1 M of HCl implies-

0.1 mol of HCl is present in 1000 mL of the solution.

∴ 0.0159 mol of HCl is present in 1000×0.0159

0.1 ml of solution

= 159 mL Of the solution

Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture

of Na2CO3 and NaHCO3, containing equimolar amounts of both.

2.7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40%

solution by mass. Calculate the mass percentage of the resulting solution.

Solution:

Given the weight of solution = 300 g

% of solution S1 = 25%

Weight of solution 2 = 400 g




% of solution S2 = 40%

The total amount of solute present in the mixture is given by,

= 300 gm × 25% + 400 gm × 40%

300 ×25

100+ 400 ×

40

100

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage (w/w) of the solute in the resulting solution,

=235

700× 100%

= 33.57%

And, mass percentage (w/w) of the solvent in the resulting solution,

= (100 − 33.57)%

= 66.43%

2.8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and

200 g of water. Calculate the molality of the solution. If the density of the

solution is 1.072 g mL−1, then what shall be the molarity of the solution?

Solution:

Given the weight of ethylene glycol = 222.6 g (solute)

Weight of water = 200 g (solvent)

We know,

The molar mass of ethylene glycol [C2H4(OH)2]

= 2 × 12 + 6 × 1 + 2 × 16

= 62 gmol−1

Molality =moles of solute

weight of solvent (kg)

Number of moles of ethylene glycol =weight of ethylene glycol

molecular weight of ethylene glycol=

222.6 g

62 gmol−1

= 3.59 mol




Therefore, molality of the solution =moles of ethylene glucol

weight of solvent (water)kg=

3.59 mol

0.200 kg=

17.95 molal

The total mass of the solution = (222.6 + 200)g = 422.6 g

Given,

The density of the solution = 1.072 g mL−1

∴ The volume of the solution =mass

density=

422.6 g

1.072 g mL−1

= 394.22 mL

= 0.3942 × 10−3L

⇒ Molarity of the solution =moles of solute

volume of solution=

3.59

0.39422×10−3 L

= 9.11 M

2.9. A sample of drinking water was found to be severely contaminated with

chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was

15 ppm (by mass):

(i). express this in per cent by mass

(ii). determine the molality of chloroform in the water sample.

Solution:

Given 15ppm by mass of solution

(i) 15 ppm (by mass) means 15 parts are present per million(106) parts of

the solution.

Therefore, percent by mass =mass of solute×100

mass of solution=

15

106 × 100%

= 1.5 × 10−5%

(ii) We know

Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol−1

Now, according to the question

15 g Of chloroform is present in 106g of the solution.

Moles of chloroform solute =weight

molecular weight=

15

119.5




The formula used molarity =moles of solute


i. e. , 15g Of chloroform is present in (106 − 15) ≈ 106 g of water.

∴ The molality of the solution =15

119.5 mol

106×10−3 kg

= 1.26 × 10−4 molal

2.10. What role does the molecular interaction play in a solution of alcohol and water?

Solution:

In pure alcohol and water, the molecules of alcohol are held tightly by strong

hydrogen bonds. The interactions between the molecules of alcohol and water are

weaker than alcohol-alcohol and water-water interactions. As a result of this,

when alcohol and water are mixed, the intermolecular interactions become weaker

and the molecules can easily escape. This lowers the boiling point of the solution.

Hydrogen bond in the ethanol molecules

Hydrogen bond in water molecule




Hydrogen bond in between Ethanol and water.

2.11. Why do gases always tend to be less soluble in liquids as the temperature is

raised?

Solution:

As temperature increases, the solubility decreases since the dissolution of gases in

liquids is an exothermic process.

The solubility of a gas decreases with the increase in temperature.

Gas + Liquid ⟶ Solution + Heat

The process is exothermic so according to Le Chatelier’s Principle the forward

reaction is exothermic thus with an increase in temperature or heat supplied the

Equilibrium will shift backwards., thereby decreasing the solubility of gases is

decreased.

2.12. State Henry’s law and mention some important applications.

Solution:

According to Henry’s law, the partial pressure of a gas in the vapour phase is

proportional to the mole fraction of the gas in the solution. If p is the partial

pressure of the gas in the vapour phase and x is the mole fraction of the gas, then

Henry’s law can be expressed as:

p ∝ x

p = KHx

Where,

KH Is Henry’s law constant.

Some important applications of Henry’s law are mentioned below.




(i) In soft drink and soda water CO2 filled bottles are under high-pressure

increase the solubility ofCO2.

Henry’s law states that the solubility of gases increases with an increase in

pressure. When a scuba diver dives deep into the sea, the high pressure

causes nitrogen to get dissolved in his blood . When the diver returns to

the surface, the solubility of nitrogen again decreases due to a decrease in

pressure upon reaching the surface. The dissolved nitrogen gas is released,

which leads to the formation of nitrogen bubbles in the blood as nitrogen

is insoluble in blood at low pressures. This process results in the blockage

of capillaries and leads to a medical condition known as ‘bends’ or

‘decompression sickness’.

(ii) The concentration of oxygen is low in the blood vessels and tissues of

people living at high altitudes. The partial pressure of oxygen at high

altitude is lower than that of the sea level. Therefore people living in high

altitudes will have less percentage of oxygen in their blood. This makes

them light-headed and weak often preventing them from thinking clearly.

2.13. The partial pressure of ethane over a solution containing 6.56 × 10−3 g of ethane

is 1 bar. If the solution contains 5.00 × 10−2 g of ethane, then what shall be the

partial pressure of the gas?

Solution:

Given partial pressure of Ethane over solution = 1 bar

Weight of ethane = 6.56 × 10−3gm

Weight of Ethane in solution = 5 × 10−2gm

We know Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1

= 30 g mol−1

∴ Number of moles present in 6.56 × 10−2g of ethane =weight (gm)

Molecular weight=

6.56×10−2

30

= 2.187 × 10−3 mol

Assume the number of moles of the solvent to be x.

According to Henry’s law,

ρ = KHx

⇒ 1 bar = KH ∙2.187 × 10−3

2.187 × 10−3 + x




⇒ 1 bar = KH2.187×10−3

x (Since x ≫ 2.187 × 10−3)

⇒ KH =x

2.187×10−3bar …. (1)

Number of moles present in 5.00 × 10−2 of ethane =5.00×10−2

30 mol

= 1.67 × 10−3 mol

Mole fraction of Ethane =moles of ethane

moles of ethane+moles of solvent=

1.67×10−3

1.67×10−3+x


ρ = KHx

=x

2.187 × 10−3×

1.67 × 10−3

(1.67 × 10−3) + x

=x

2.187×10−3 ×1.67×10−3

x (Since, x ≫ 1.67 × 10−3)

= 0.764 bar

2.14. What is meant by positive and negative deviations from Raoult's law and how is

the sign of ∆mixH related to positive and negative deviations from Raoult's law?

Solution:

According to Raoult’s law, the partial vapour pressure of each volatile component

in any solution is directly proportional to its mole fraction.

Present in the solution p1 ∝ x1

And p1 = p1°x1

Where p° = vapour pressure of pure component at all the same temperature

p = partial p vapour pressure of component

x = mole fraction of volatile component

The solutions that obey Raoult’s law at each and every concentration of the

solution are termed as ideal solutions.

The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour

pressures either higher or lower than that predicted by Raoult’s law. If the vapour

pressure is higher, then the solution is said to exhibit positive deviation.

Vapour pressure of a two-component solution showing positive deviation from

Raoult’s law: ΔS > 0 ; ΔG > 0 ; Vp(mixture) > 0 ; ΔH < 0




A solution that shows negative deviation: If vapour pressure is lower than the

predicted value, then the solution is said to exhibit negative deviation from

Raoult’s law.

Vapour pressure of a two-component solution showing negative deviation from

Raoult’s law:

In the case of an ideal solution, the enthalpy of the mixing of the pure components

for forming the solution is zero. ΔS > 0 ; ΔG < 0 ;Vmix < 0 ; ΔH > 0

In solutions showing negative deviation, heat gets evolved from the solution.

2.15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at

the normal boiling point of the solvent. What is the molar mass of the solute?

Solution:

Aqueous solution of non-volatile solute = 2%

Here,

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar

We know Vapour pressure of pure water (solvent) at normal boiling point (p1o) =

1.013 bar

Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) mass of solution – mass of solute = 100 − 2




The molar mass of solvent (water), (M1) = 2 × 1 + 16 = 18 g mol−1

According to Raoult’s law,

Let n1& n2 are number of moles of solvent and solute

Mole of solute =(n1)

moles of solution (n2)

Moles of solvent (n2) =weight

molecular weight=

98

18

Moles of solute (n1) =weight

molecule weight=

1

M1

p10 − p1

p10 =

w1 × M2

M1 × w2

⇒1.013 − 1.004

1.013=

2 × 18

M2 × 98

⇒0.009

1.013=

2 × 18

M1 × 98

⇒ M1 =1.013 × 2 × 18

0.009 × 98

= 41.35 g mol−1

2.16. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the

two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the

vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Solution:

Given,

Vapour pressure of heptane (p10) = 105.2 kPa

Weight of heptane = 26.0 g

Weight of octane = 35 g

Vapour pressure of octane (p20) = 46.8 kPa

We know that,

Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1

= 100 g mol−1

∴ Number of moles of heptane =weight of heptane

molecular weight of heptane=

26

100 mol




= 0.26 mol

Molar mass of octane (C8H18) = 8 × 12 + 18 × 1

= 114 g mol−1

∴ Number of moles of octane =weight

molecular weight=

35

114 mol

= 0.31 mol

Mole fraction of heptane, (x1) =moles of heptane

moles of heptane+ moles of octane=

0.26

0.26+0.31

0.456

And, mole fraction of octane, (x2) = 1 − 0.456

= 0.544

Now, the partial pressure of heptane, = moles fraction of heptane × vapour

pressure of heptane(p1°);

(p1) = x1p10

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane, = mole fraction of octane × vapour pressure of

octane(p2°);

(p2) = x2p20

= 0.544 × 46.8

=25.4592 kPa

2.17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure

of 1 molal solution of a non-volatile solute in it.

Solution:

Given vapour pressure of water 12.3 kPa

Molality of solution = 1 (molal)

1 Molal solution means one mol of the solute is present in 1000 g of the solvent

(water).

Mole of solute = 1

Weight of solvent = 1000 g




Molar mass of water = 18 g mol−1

∴ Number of moles present in 1000 g of water =1000

18

= 55.56 mol

Therefore, mole fraction of the solute present in the solution (x2) =moles of solute

moles of solute+moles of solvent=

1

1+55.56= 0.0177

Relative vapour pressure = mole fraction of solute (x2)

p1° = Vapour pressure of solvent

p1 = Vapour pressure of solution

Formula, p1

0−p1

p10 = x2 ⇒

12.3−p1

12.3= 0.0177 ⇒ 12.3 − p1 = 0.217 ⇒ p1 =

12.0823

= 12.08 kPa (Approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

2.18. Calculate the mass of a non-volatile solute (molar mass40 g mol−1) which should

be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution:

Given molar mass of non-volatile solute = 40 g/mol

Solvent = octane


Let the vapour pressure of pure octane be p10.

Then, from the question the vapour pressure of the octane after dissolving the

non-volatile solute = 80

100p1

0 = 0.8p10.

Assume weight of solute = w2

Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1

= 114 g mol−1

Formula use

vapour pressure of solvent − vapour pressure of solutions

vapour pressure of solvent

=moles of solute

moles of solvent




Moles of solute =w2

40

Moles of solvent =40

114

p10 − p1

p10 =

w2 × M1

M2 × w1

⇒p1

0 − 0.8p10

p10 =

w2 × 114

40 × 114

⇒0.2p1

0

p10 =

w2

40

⇒ 0.2 =w2

40

⇒ w2 = 8 g

Hence, the mass required of the solute is 8 g.

2.19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a

vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the

solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(i) molar mass of the solute

(ii) vapour pressure of water at 298 K.

Solution:

Given,

Weight of non-volatile solute = 30 g

Weight of H2O (solvent) = 90 g

Added amount of water = 18 g in solution

(i) Let the molar mass of the solute = M g mol−1

Number of moles of solvent (water), (n1) =weight of water

molecular weight of water=

90 g

18 g mol−1= 5mol

And, the no. of moles of solute, (n2) =weight of solute

molecular weight of solute=

30 g

M mol−1=

30

Mmol

Formula:




vapour pressure of solvent (p1°) − vapour pressure (p1)

vapour pressure of solvent (p1°)

= mole fraction of solute(x)

mole fraction of solute =moles of solute

moles of solute + mole of solvent

=

30M

5 +30M

p1 = 2.8 kPa

p10 − p1

p10 =

n2

n1 + n2

⇒p1

0 − 2.8

p10 =

30M

5 +30M

⇒ 1 −2.8

p10 =

30M

5M + 30M

⇒ 1 −2.8

p10 =

30

5M + 30

⇒2.8

p10 = 1 −

30

5M + 30

⇒2.8

p10 =

5M

5M + 30

⇒

2.8

p10

2.8=

5M+30

5M…………………. (i)

After the addition of 18 g of water:

Total weight of solvent (water) = 90 + 18

Moles of water (n1) =weight of water

molecular weight of water=

90+18 g

18= 6mol

Moles of solute =moles of solute (n2)

moles of solute(n1)+mole of solvent(n2)

p1 = 2.9kPa

Formula:




p1° = Vapour pressure of solvent

p1 = Vapour pressure of solution = 2.9 × kPa

Moles of solute =weight

molecular weight=

30

M

p10 − p1

p10 =

n2

n1 + n2

⇒p1

0 − 2.9

p10 =

30/M

6 +30M

⇒ 1 −2.9

p10 =

30/M

6M + 30M

⇒ 1 −2.9

p10 =

30

6M + 30

⇒2.9

p10 =

6M + 30 − 30

6M + 30

⇒2.9

p10 =

6M

6M + 30

⇒p1

0

2.9=

6M+30

6M …………………… (ii)

Dividing equation (i) by (ii), we have:

2.9

2.8=

5M + 305M

6M + 306M

⇒2.9

2.8=

(5M + 30) × 6M

5M(6M + 30)

⇒ 2.9 × 5 × (6M + 30) = 2.8 × 6 × (5M + 30)

⇒ 87M + 435 = 84M + 504

⇒ 3M = 69

⇒ M = 23u

Therefore, the molar mass of the solute is 23 g mol−1.

(ii) Substituting the value ofM in eqn (i), we get:




p10

2.8=

5 × 23 × 30

5 × 23

p10

2.8= 145/115

⇒ p10 = 3.53

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

2.20. A 5% solution (by mass) of cane sugar in water has a freezing point of 271K.

Calculate the freezing point of 5% glucose in water if the freezing point of pure

water is 273.15 K.

Solution:

Given:

Freezing point can sugar in of water = 271 K

Freezing point of water = 0°C = 273.15K

ΔTf = Freezing point of solvent – freezing point of solution

Here, ∆Tf = (273.15 − 271)K

= 2.15 K

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16

= 342 g mol−1

5% solution by mass of cane sugar in H2O means 5 g of cane sugar is present in

(100 − 5)g = 95g of water.

Now, number of moles of cane sugar =weight

molecular weight=

5

342 mol

= 0.0146 mol

Therefore, molality of the solution, (m) =moles of solute

weight of solvent (kg)=

0.0146 mol

0.095 kg

= 0.1537 mol kg−1

Formula relative lowering vapour pressure ∝ molality

Relative lowering vapour pressure = kg × molality

∆Tf = Kf × m

⇒ Kf =∆Tf

m




=2.15K

0.1537 mol kg−1

= 13.99 K kg mol−1

(ii) Given freezing point of glucose in water =?

Freezing point of water (solvent) = 273.15



= 0.95 g

We know Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol−1

5% Glucose in water 5 g of glucose is present in (100 − 5)g = 95 g of

water.

∴ Number of moles of glucose =weight

molecular weight=

5

180 mol

= 0.0278 mol

Therefore, molality of the solution, (m) =moles of solute

weight solvent in kg=

0.0278 mol

0.095 kg

= 0.2926 mol kg−1

Formula

⇒ Relative lowering in freezing point ∝ molality

⇒ Relative lowering in freezing point = kg × molality

∆Tf = Kf × m

= 13.99 K kg mol−1 × 0.2926 mol kg−1

ΔTf = 4.09 K

Hence, the freezing point of 5% glucose solution ΔTf = freezing point of

solvent – freezing point of solution

⇒ (273.15 − 4.09)K = 269.06 K.

2.21. Two elements A and B form compounds having formula AB2 and AB4. When

dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by

2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for

benzene is 5.1 K kg mol−1. Calculate atomic masses of A and B.




Solution:

Given the formula of compound AB2 and AB3

Weight of Benzene = 20 gm

Weight of AB2 = 1 g ; (ΔTf)AB2= 3.5

Weight of AB4 = 1g ; (ΔTf)AB4= 1.3

Formula ΔTf = kf × molality × i

∴ Molality =moles of solute×100


i = Van't Hoff factor

M = i1000 × w2 × kf

∆Tf × w1

Then, MAB2=

1000×1×5.1

2.3×20

= 110.87 g mol−1

MAB4=

1000 × 1 × 5.1

1.3 × 20

= 196.15 g mol−1

Now, we have the molar mass of AB2 and AB4 as 110.87 g mol−1 and

196.15 g mol−1 respectively.

Assume let the atomic masses of A and B x and y respectively, and they are non-

electrolyte.

Now, we can write:

x + 2y = 110.87 ……………(i)

x + 4y = 196.15 .………….(ii)

Subtracting equation (i) from (ii), we have

2y = 85.28

y = 42.64

Putting the value of ′y′ in equation (1), we have

x + 2 × 42.64 = 110.87

x = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u, respectively.




2.22. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure

of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same

temperature, what would be its concentration?

Solution:

Given,

i = Van't Hoff factor ⇒ for glucose (i) = 1

Concentration of glucose =36

180= M

∴ Molecular weight of glucose = 180

Temperature (T1) = 300 K

Osmotic pressure (π) = 1.52 bar

Gas constant R = 0.083 bar LK−1mol−1

Osmotic pressure(π2) = 1.52

T2 = 300 K

Formula: π = iCRT

⇒ π = CRT ⇒ C =π

RT

⇒π1

C1=

π2

C2⇒

4.98

0.2=

1.52

C2

⇒ C2 =1.52 × 0.2

4.98

= 0.61 M

As the volume of the solution is 1L, the concentration of the solution would be

0.061 M.

2.23. Suggest the most important type of intermolecular attractive interaction in the

following pairs.

(i) n-hexane and n-octane

(ii) I2 And CCl4

(iii) NaClO4 And water

(iv) Methanol and acetone

(v) Acetonitrile (CH3CN) and acetone (C3H6O).




Solution:

(i) n-hexane and n-octane both are gases, so they have Van der Waals forces

of attraction.

(ii) I2 And CCl4 both are covalent compound so they have Van der Waals

forces of attraction.

(iii) Na+ClO4− is an ionic compound and H2O is polar compound, so force

between then is Ion-dipole interaction.

(iv) Method CH2OH and acetone both are polar compounds,

so they have Dipole-dipole interaction.

(v) Acetonitrile (CH3 − CN−δ) and acetone both are polar

compounds, so they have Dipole-dipole interaction.

2.24. Based on solute-solvent interactions, arrange the following in order of increasing

solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

Solution:

n −Octane is a non-polar solvent. Therefore, the solubility of non-polar solute is

more than that of a polar solute in the n −octane.

The order of increasing polarity is:

Cyclohexane< CH3CN < CH3OH < KCl (ionic)

∴ ∈ N of O > N so more polar

Therefore, the order of increasing solubility in nonpolar solvent octane is

KCl < CH3OH < CH3CN < Cyclohexane

2.25. Amongst the following compounds, identify which are insoluble, partially soluble

and highly soluble in water?

(i) Phenol

(ii). Toluene

(iii). formic acid

(iv). ethylene glycol

(v). chloroform




(vi). pentanol.

Solution:

(i) Phenol (C6H5OH) has the polar group - OH and non-polar group - C6H5. Thus, phenol is partially soluble in water due to H-bonding, H-bond is

weak due to crowding.

(ii) Toluene (C6H5 − CH3) has no polar groups. Thus, toluene is insoluble in

water.

(iii) Formic acid (HCOOH) has the polar group - OH and thus can form H-bond

with water. Therefore formic acid is highly soluble in water.

(iv) Ethylene glycol has polar – OH group and can form

H-bond. Thus, it is highly soluble in water.

(v) Chloroform is insoluble in water because CCl4 is non-polar

(vi) Pentanol (C5H11OH) has polar (−OH) group, but it also contains a bulky

non-polar - C5H11 group. Thus, pentanol is partially soluble in water due

to poor H-bonding.

2.26. If the density of some lake water is 1.25g mL−1 and contains 92 g of Na+ ions per

kg of water, calculate the molality of Na+ ions in the lake.

Solution:

Given the density of lake water= 1.25gm

ml

Weight of Na+ ion = 92 gm

Weight of solvent (water) = 1 kg

Moles =weight

Atomic weight of Na

Number of moles present in 92 g if Na+ ions =92 g

23 g mol−1

= 4 molal

Formula : molality =moles of solute

weight of solvent (kg)molal




Therefore, molality of Na+ ions in the lake =4 mol

1 kg

= 4 m

2.27. If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of

CuS in aqueous solution.

Solution:

Given, Solubility product of CuS, Ksp = 6 × 10−16

Assume Let s be the solubility of CuS in mol L−1.

Now, Ksp = [Cu2+][S2−]

= s × s

Ksp = s2

Then, we have, Ksp = s2 = 6 × 10−16

⇒ s = √6 × 10−16

= 2.45 × 10−8mol L−1

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 ×

10−8 mol L−1.

2.28. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when

6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Solution:

Given the weight of aspirin (C9H8O4) = 6.5 gm

6.5 g of C9H8O4 is dissolved in 450 g CH3CN.

Then, the total mass of the solution (weight of C9H8O4 + weight CH3CN) =(6.5 + 450)g

= 456.5 g

Therefore, mass percentage of C9H8O4 =weight of C9H8O4

total weight of solution

=weight of C9H8O4

weight of C9H8O4 + weight of CH3CN=

6.5

456.5× 100%

= 1.424%




2.29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal

symptoms in narcotic users. The dose of nalorphene generally given is 1.5 mg.

Calculate the mass of 1.5 × 10−3 m aqueous solution required for the above dose.

Solution:

Molality 1.5 × 10−3m

Dose of nalorphene given = 1.3 × 10−3gm

The molar mass of nalorphine (C19H21NO3) is given as:

19 × 12 + 21 × 1 + 1 × 14 + 3 × 16 = 311 g mol−1

In 1.5 × 10−3m aqueous solution of nalorphine,

i.e., 1 kg (1000 g) of water contains 1.5 × 10−3 mol = 1.5 × 10−3 × 311 g

= 0.46665 g

Therefore, the total mass of the solution = (1000 + 0.4665) g

= 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphine is

1000.4665 g.

Therefore, the mass of the solution containing 1.5 mg of nalorphine =1000.4665×1.5×10−3

0.4665 g

= 3.22 g

Hence, the mass of aqueous solution required is 3.22 gm.

2.30. Calculate the amount of benzoic acid (C6H5COOH) required for preparing

250 mL of 0.15 M solution in methanol.

Solution:

Given

Vmethanol of solution = 250 ml

Mmethanol of solution = 0.15 M

0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mole of benzoic acid

Therefore, 250 mL of solution contains =0.15×250

1000 mol of benzoic acid

= 0.0375 mol of benzoic acid




The molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16

= 122 g mol−1

Hence, required benzoic acid = mole of benzoic acid × molecular weight of

benzoic acid = 0.0375 mol × 122 g mol−1.

= 4.575 g

2.31. The depression in freezing point of water observed for the same amount of acetic

acid, trichloroacetic acid and trifluoroacetic acid increases in the order given

above. Explain briefly.

Solution:

(ΔTf) ⇒ CH3 − COOH < Cl3CCOOH < F3CCOOH

Among H, Cl, and F, H is least electronegative while F is most electronegative in

Pauling scale =4.

Cl = 3.3

O = 3.5

H = 2.5

Then, F can withdraw electrons towards itself more than Cl and H. Thus,

trifluoroacetic acid can easily lose H+ ions i. e., trifluoroacetic acid ionizes to the

largest extent Now, the more ions produced, because as acidic character increase

H+ ion and anion availability increases, so I factor increases. The greater the

depression of the freezing point. Hence, the depression in the freezing point

increases in the order:

ΔTf ∝ i

i ∝ acidic strength

Acetic acid < trichloroacetic acid < trifluoroacetic acid

2.32. Calculate the depression in the freezing point of water when 10 g of

CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3, Kf =

1.86 K kg mol−1.

Solution:




Given

Weight of CH3 − CH2 − CHClCCOH = 10gm

Weight of water = 250 gm

Ka = 1.4 × 10−3

kf = 1.86 k kg

ml

We know Molar mass of CH3CH2CHClCOOH = 122.5 g mol−1

∴ No. of moles present in 10 g of

CH3CH2CHClCOOH =weight

molecular weight=

10 g

122.5 g mol−1

= 0.0816 mol

It is given that 10 g of CH3CH2CHClCOOH is added to 250 g of water.

∴ Molality of the solution(m)

=moles of solute × 1000

weight of solvent =

0.0186

250× 1000 molal or mol kg−1

= 0.3264 mol kg−1

Let α be the degree of dissociation of CH3CH2CHClCOOH.

CH3CH2CHClCOOH undergoes dissociation according to the following equation:

CH3CH2CHClCOOH ↔ CH3CH2CHClCOO− +H+

Initial cone. C mol L−1 0 0

At equilibrium C(l − α) Cα Cα

∴ Kα =Cα. Cα

C(l − α)

=Cα2

l − α

Since α is very small with respect to 1, 1 − α ≈ 1

Now, Kα =Cα2

1

⇒ Kα = Cα2




⇒ α = √Kα

C

= √1.4 × 10−3

0.3264 (∵ Kα = 1.4 × 10−3)

= 0.0655

Again,

CH3CH2CHClCOOH ↔ CH3CH2CHClCOO− +H+

Initial moles 1 0 0

At equilibrium l − α α α

Total moles of equilibrium = 1 − α + α + α = 1 + α

⇒ i =moles dissociated

initial moles

i =1 + α

1

= 1 + α

= 1 + 0.0655

i = 1.0655

Hence, the depression in the freezing point of water is given as:

∆Tf = iKfm

∆Tf = 1.0655 × 1.86K kg mol−1 × 0.3264 mol kg−1

= 0.65 K

2.33. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the

freezing point of water observed is 1.0o C. Calculate the van’t Hoff factor and

dissociation constant of fluoroacetic acid.

Solution:

Given that:

Weight of water (w1) = 500 g

Weight of CH2FCOOH (w2) = 19.5 g




Kf = 1.86 K kg mol−1

∆Tf = 1K

Formula

⇒ ΔTf = kg × m

Molality =moles of solute×1000


M =

w2

M2w1

1000

=w2

M2×

1000

w1

M2 =Kf × w2 × 1000

∆Tf × w1

=1.86 K kg mol−1 × 19.5 g × 1000g kg−1

500 g × 1 K

M2 = 72.54 g mol−1

Therefore, the observed molar mass of CH2FCOOH, (M2)obs = 72.54g mol

Molar mass of CH2FCOOH is:

(M2)cal = 14 + 19 + 12 + 16 + 16 + 1

= 78 g mol−1

Therefore, van’t Hoff factor, i =(M2)cal

(M2)obs

=78 g mol−1

72.54 g mol−1

= 1.0753

Let α be the degree of dissociation of CH2FCOOH

CH2FCOOH ↔ CH2FCOO− +H+

Initial conc. C mol L−1 0 0

At

equilibrium

C(1 − α) Cα Cα Total = C(1 + α)

α =degree of dissociation

∴ i =concentration after dissociation

initial concentration

C(1 + α)

C




⇒ i = 1 + α

⇒ α = i − 1

= 1.0753 − 1

= 0.0753

Now the value of Ka is given as:

Ka =[CH2FCOO−][H+]

[CH2FCOOH]

=Cα. Cα

C(1 − α)

=Cα2

1 − α

Taking the volume of the solution as 500 mL, we have the concentration:

C =

19.578

500× 1000M

= 0.5M

Therefore, Ka =Cα2

1−α

=0.5 × (0.0753)2

1 − 0.0753

=0.5 × 0.00567

0.9247

= 0.00307 (approximately)

= 3.07 × 10−3

2.34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour

pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Solution:

Given,

Vapour pressure of water, p10 = 17.535 mm of Hg

Temperature = 293 K

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g




We know that,

The molar mass of glucose C6H12O6, (M2) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol−1

The molar mass of water, (M1) = 18 g mol−1

Then, the number of moles of glucose,

(n2) =weight

molecular weight=

25

180 g mol−1 = 0.139 mol

Mole fraction of solute =n1

n1+n2

And, the number of moles of water,

n1 =weight

molecular weight=

450 g

18 g mol−1= 25 mol

We know that,

p10−p1

p10 = mole fraction of solute =

n2

n2+n1

⇒17.535 − p1

17.535=

0.139

0.139 + 25

⇒ 17.535 − p1 =0.139 × 17.535

25.139

⇒ 17.535 − p1 = 0.097

⇒ p1 = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.

2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 ×

105 mm Hg. Calculate the solubility of methane in benzene at 298 K under

760 mm Hg.

Solution:

Given

p = 760 mm Hg

kH = 4.27 × 105 mm Hg


Partial pressure = k4 × mole fraction methane in benzene




p = kHx

⇒ x =p

kH

=760 mm Hg

4.27 × 105 mm Hg

= 177.99 × 10−5

x = 178 × 10−5 (approximately)

Hence, the mole fraction of methane in benzene is 178 × 10−5.

2.36. 100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B

(molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be

500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in

the solution if the total vapour pressure of the solution is 475 Torr.

Solution:

Given the weight of A(liquid) = 100 gm

Molecular weight of A = 140 gm/mole

Weight of liquid B = 1000 gm

Molecular weight of B = 180 gm/mole

Vapour pressure of pure liquid B = 500 torr.

Vapour pressure of pure liquid A =?

Total vapour pressure = 475 torr.

Number of moles of liquid B, (nB) =weight

molecular weight=

1000

180 mol

= 5.556 mol

Then, the mole fraction of A, (xA) =weight

molecular weight=

nA

nA+nB

=0.714

0.714 + 5.556

= 0.114

And, mole fraction of B, xB = 1 − 0.114

= 0.886

Vapour pressure of pure liquid B, (pB0 ) = 500 torr




Therefore, the vapour pressure of liquid B in the solution,

pB = pB0 xB

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, (ptotal) = 475 torr

∴ PA + PB = Ptotal

∴ Vapour pressure of liquid A in the solution,

(pA) = ptotal − pB

= 475 − 443

= 32 torr

Now,

pA = pA0 xA

⇒ pA0 =

pA

xA

=32

0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

2.37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and

632.8 mm Hg respectively. Assuming that they form an ideal solution over the

entire range of composition, plot ptotal, pchloroform, and pacetone as a function of

xacetone. The experimental data observed for different compositions of the

mixture is:

Plot this data also on the same graph paper. Indicate whether it has a positive

deviation or negative deviation from the ideal solution.

100

× xacetone

0 11.8 23.4 36.0 50.8 58.2 64.5 72.1

pacetone

/ mm Hg

0 54.9 110.1 202.4 322.7 405.9 454.1 521.1

pchloroform

/ mm Hg

6

3

2

5

4

8

4

6

9

3

5

9

2

5

7

1

9

3

1

6

1

1

2

0




.

8

.

1

.

4

.

7

.

7

.

6

.

2

.

7

ptotal(mm Hg) 6

3

2

.

8

6

0

3

.

0

5

7

9

.

5

5

6

2

.

1

5

8

0

.

4

5

9

9

.

5

6

1

5

.

3

6

4

1

.

8

Solution:

From the question, we have the following data

100

× xacetone

0 11.

8

23.

4

36.

0

50.

8

58.

2

64.

5

72.

1

pacetone

/ mm Hg

0 54.

9

110

.1

202

.4

322

.7

405

.9

454

.1

521

.1

pchloroform

/ mm Hg

632

.8

548

.1

469

.4

359

.7

257

.7

193

.6

161

.2

120

.7

ptotal(mm Hg) 632

.8

603

.0

579

.5

562

.1

580

.4

599

.5

615

.3

641

.8

The plot of ptotal slopes downwards. Therefore, the solution is said to show a

negative deviation from Raoult’s law.

2.38. Benzene and toluene form an ideal solution over the entire range of composition.

The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and

32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour

phase if 80 g of benzene is mixed with 100 g of toluene.

Solution:




Given,

Vapour pressure of pure benzene = 50.71mm of Hg

Vapour pressure of toluene = 32.06 mm of Hg

Weight of benzene in vapour phase = 80 g

Weight of toluene in vapour phase = 100 g

We know,Molar mass of benzene (C6H6) = 6 × 12 + 6 × 1

= 78 g mol−1

Molar mass of toluene (C6H5CH3) = 7 × 12 + 8 × 1

= 92 g mol−1

Now, no. of moles present in 80 g of benzene

=weight

molecular weight=

80

78 mol = 1.026 mol

And, no. of moles present in 100 g of toluene

=weight

molecular weight=

100

92mol = 1.087 mol

∴ Mole fraction of benzene, (xb) =1.026

1.026+1.087= 0.486

∴ xB + xt = 1 so xt = 1 − xB

And, mole fraction of toluene, (xt) = 1 − 0.486 = 0.514

Vapour pressure of pure benzene, (pb0) = 50.71 mm Hg

And, vapour pressure of pure toluene, (pt0) = 32.06 mm Hg

Therefore, partial vapour pressure of benzene, (pb) = xb × pb

= 0.486 × 50.71

= 24.645 mm Hg

And, partial vapour pressure of toluene, (pt) = xt × pt

= 0.514 × 32.06

= 16.479 mm Hg

∴ the mole fraction of C6H6 in vapour phase is given to be




=pb

pb + pt=

24.645

24.645 + 16.479

=24.645

41.124

= 0.599

= 0.6

2.39. The air is a mixture of a number of gases. The major components are oxygen and

nitrogen with an approximate proportion of 20% is to 79% by volume at 298 K.

The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the

Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and

6.51 × 107 mm respectively, calculate the composition of these gases in water.

Solution:

Given

v ∝ n;

nO2= 20 and nN2

= 79

Total moles = 100

Percentage of oxygen (O2) in air = 20%

Percentage of nitrogen (N2) in atmosphere = 79 %

It is also given that water is in equilibrium with air at a total pressure of 10 atm,

(PT) = (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen,

(pO2) = mole fraction of O2 × Pt =

moles of O2

total mass× Pt =

20

100× 7600 mm Hg

= 1520 mm Hg

Partial pressure of nitrogen,

(pN2) = mole fraction of N2 × Pt =

moles of N2

total mass=

79

100× 7600 mm Hg

= 6004 mm Hg

Now, according to Henry’s law:

p = KH. x




For oxygen:

pO2= KH. xO2

⇒ xO2=

pO2

KH

=1520 mm Hg

3.30×107 mm Hg (Given KH = 3.30 × 107 mm Hg)

= 4.61 × 10−5

For nitrogen:

pN2= KH ∙ xN2

⇒ xN2=

pN2

KH

=6004 mm Hg

6.51 × 107 mm Hg

Therefore, the mole fraction of nitrogen is 4.61 × 10−5 and of oxygen is 9.22 ×

10−5.

2.40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litres of water such

that its osmotic pressure is 0.75 atm at 27o C.

Solution:

Given

Formula π = iCRT ∴ C =moles

volume

We know that,

Osmotic pressure π = 0.75 atm

(Volume) V = 2.5L

(Van't Hoff factor) i = 2.47

Temperature (T) = (27 + 273)K = 300K

We know

R = 0.0821 L atm K−1mol−1

The molecular weight of CaCl2 = 1 × 40 + 2 × 35.5

= 111g mol−1




π = in

VRT

⇒ π = iw

MVRT

⇒ w =πMV

iRT

Therefore, w =0.75×111×2.5

2.47×0.0821×300

= 3.42 g

Hence, the required amount of CaCl2 is 3.42 g.

2.41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of

K2SO4 in 2 litres of water at 25oC, assuming that it is completely dissociated.

Solution:

Given,

weight ofK2SO4 = 25 mg = 0.025 g

Volume of solution = 2 L

T = 25oC = (25 + 273)K = 298 K

We know that:

R = 0.0821 L atm K−1mol−1

M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol−1

Applying the following relation,

π = iCRT ∴ C =mole

volume

π = in

vRT

When K2SO4 is dissolved in water, K+ and SO42− ions are produced.

K2SO4 ⟶ 2K+ + SO42−

Total number of ions produced = 3 (100% dissociation)

∴ i = 3

π = in

vRT




⇒ iw

M

1

vRT

= 3 ×0.025

174×

1

2× 0.0821 × 298

= 5.27 × 10−3 atm

⧫ ⧫ ⧫


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